Question

我使用brian2来运行神经网络模拟。为了在每次模拟过程中记录数据，我创建了brian2 SpikeMonitor类的多个实例。我想将这些监视器存储在一个dict中，使用dict理解创建。

作为测试，我在交互式会话中执行以下操作：

In [1]: import brian2

In [2]: pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)

In [3]: record_pops = ['pe_mt']

In [4]: {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in record_pops}
Out[4]: {'mon_pe_mt': <SpikeMonitor, recording spikemonitor>}

一切看起来都很棒。但现在当我将此代码移动到以下函数

时

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    return {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in
            record_pops}

并调用它，我收到以下错误

In [9]: tests.test_record()
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-9-4d3d585b2c97> in <module>()
----> 1 tests.test_record()

/home/daniel/Science/dopa_net/brian/ardid/tests.py in test_record()
     61     record_pops = ['pe_mt']
     62     return {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in
---> 63                 record_pops}
     64     # DEBUG ###################
     65     #monitors = utils.record(['pe_mt'], 'spikes', None, None, pe_mt, None, None)

/home/daniel/Science/dopa_net/brian/ardid/tests.py in <dictcomp>((pop,))
     60     # DEBUG ###################
     61     record_pops = ['pe_mt']
---> 62     return {'mon_' + pop: brian2.SpikeMonitor(eval(pop)) for pop in
     63                 record_pops}
     64     # DEBUG ###################

/home/daniel/Science/dopa_net/brian/ardid/tests.py in <module>()

NameError: name 'pe_mt' is not defined

这里发生了什么？＆＃39; pe_mt＆＃39;在函数中定义了。

请注意，如果我将dict理解更改为列表理解，如

return [brian2.SpikeMonitor(eval(pop)) for pop in record_pops]

没有错误提出！我得到了SpikeMonitor个对象的列表，这些对象已经适当定义。

现已删除的答案表明我使用locals()[pop]代替eval(pop)。请注意，这会产生一个等效错误：

In [20]: tests.test_record()
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-20-4d3d585b2c97> in <module>()
----> 1 tests.test_record()

/home/daniel/Science/dopa_net/brian/ardid/tests.py in test_record()
     61     record_pops = ['pe_mt']
     62     return {'mon_' + pop: brian2.SpikeMonitor(locals()[pop]) for pop in
---> 63                 record_pops}
     64     # DEBUG ###################
     65     #monitors = utils.record(['pe_mt'], 'spikes', None, None, pe_mt, None, None)

/home/daniel/Science/dopa_net/brian/ardid/tests.py in <dictcomp>((pop,))
     60     # DEBUG ###################
     61     record_pops = ['pe_mt']
---> 62     return {'mon_' + pop: brian2.SpikeMonitor(locals()[pop]) for pop in
     63                 record_pops}
     64     # DEBUG ###################

KeyError: 'pe_mt'

Answer 1

一：忘记eval，因为如果传递给它的字符串是表达式或函数调用而不是标识符，它可能会导致意外事情发生。如果确实需要按名称获取本地变量，则可以使用locals()[name]干净利落地完成。

文档：locals

二：所有的理解和生成器表达式（除了python 2.x中的列表推导除外）have their own namespace，因此理解中的locals()将引用那个 - 没有你的变量的那个。对于captures your local variables by default：

的eval也是如此

如果省略locals字典，则默认为globals字典。如果省略两个字典，则表达式在调用eval（）的环境中执行。

你可以通过早点获得它们来解决这个问题：

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    groups = locals()    
    return {'mon_' + pop: brian2.SpikeMonitor(eval(pop, globals(), groups)) for pop in record_pops}
    # or better
    return {'mon_' + pop: brian2.SpikeMonitor(groups[pop]) for pop in record_pops}

或更常规，没有locals：

def test_record():
    groups = {
        "pe_mt": brian2.PoissonGroup(1, 100 * brian2.Hz),
    }
    return {'mon_' + key: brian2.SpikeMonitor(value) for key, value in groups.iteritems()}

Answer 2

不推荐的解决方法：

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    my_loc = locals()
    return {'mon_' + pop: brian2.SpikeMonitor(eval(my_loc[pop])) for pop in
            record_pops}

或使用普通循环来构建你的词典：

def test_record():
    pe_mt = brian2.PoissonGroup(1, 100 * brian2.Hz)
    record_pops = ['pe_mt']
    d = {}
    for pop in record_pops:
        d['mon_' + pop] = brian2.SpikeMonitor(locals()[pop]))
    return d

或者只是使用dict来保存对象：

def test_record():
    d = {"pe_mt":brian2.PoissonGroup(1, 100 * brian2.Hz)}
    record_pops = ['pe_mt']
    return {'mon_' + pop: brian2.SpikeMonitor(d[pop]) for pop in record_pops}

Dict理解产生看似无根据的NameError

2 个答案: